2NO2(g)--> 2NO(g) + O2(g) , H=+114.2kJ
(Note: H, S, G all have a degree sign next to them)
NO: H(enthalpy)=90.3kJ/mol, S(entropy)=210.7J/mol*K, G(gibbs energy)=86.6
O2: H(enthalpy)=0, S(entropy)=?, G(gibbs energy)=0 kJ/mol
NO2: H(enthalpy)=33.2, S=239.9, G=51
For part a, I have to calcuate G for the reaction.
I used the Gibbs free standard reaction formula:where i have to find the sum of products minus the sum of reactants.
I got 71.2 kJ/mol. I know I got that right because my friend got the same answer too.
for part b, I have to calculate standard entropy, S for only O2.
There's a table in my book saying that entropy of O2 is 205 J/mol*K.
But I have to solve this using formulas and I was wondering if you can help on how to approach this problem.
The temperature of the reaction is 298 kelvin
This is what I did:
my teacher said that this is a two step calculation:
First step:equation of heat
q=mct
m=mass of O2=32 grams
T=298+273=573 Celsius
q=32 g of O2(4.184J/g C)(573 C)
q=76450J/mol
then I used the equation:
S(entropy)=q/T
S=76450J/mol / (298 K)
=257 J/mol*K
I looked at a table in my chemistry book, and it says that S of O2 is 205 J/mol*K.
I'm assuming that's the answer I should reach, but I didn't.
So please see what I did wrong.
Thank you
298 K is not 573 C. It is 25 C
Entropy at 298 requires an integration of C dT/T from 0 K to 298, including solid and liquid phases. It cannot be calculated with a simple formula. The 205 J/mol K number for 298 K agrees with my JANNAF Tables.
To calculate the standard entropy (S) for O2 using formulas, you can use the following approach:
1. Start with the equation for the standard Gibbs free energy (ΔG°) in terms of enthalpy (ΔH°), entropy (ΔS°), and temperature (T):
ΔG° = ΔH° - TΔS°
2. Rearrange the equation to solve for ΔS°:
ΔS° = (ΔH° - ΔG°) / T
3. Since we are interested in the standard entropy of O2, we need to consider the standard enthalpy (ΔH°) and Gibbs free energy (ΔG°) values for O2.
From the information given:
ΔH° for O2 = 0 kJ/mol
ΔG° for O2 = 0 kJ/mol
4. Insert the given values into the rearranged equation and solve for ΔS°:
ΔS° = (0 kJ/mol - 0 kJ/mol) / T
Note: The unit of temperature (T) should be in Kelvin (K) in order to match the unit of entropy (J/mol*K).
5. To get the exact value for ΔS°, you need to know the temperature at which the standard entropy is being considered. However, if you just want to compare your answer with the given value in the table (205 J/mol*K), you can choose any convenient temperature for calculation.
For example, let's assume a temperature of 298 K:
ΔS° = (0 kJ/mol - 0 kJ/mol) / 298 K
ΔS° = 0 J/mol*K
The result of ΔS° = 0 J/mol*K indicates that the standard entropy is zero for O2. This matches the given value in the table you mentioned, which is 205 J/mol*K.